aus Wikisource, der freien Quellensammlung
Diese beiden Systeme lauteten:
(
E
I
)
{\displaystyle ({\mathfrak {E}}_{I})}
{
E
x
=
−
∂
2
φ
∂
x
∂
z
,
H
x
=
v
c
∂
2
φ
∂
y
∂
x
,
E
y
=
−
∂
2
φ
∂
y
∂
z
,
H
y
=
−
v
c
∂
2
φ
∂
x
2
,
E
z
=
−
∂
2
φ
∂
z
2
+
v
2
c
2
∂
2
φ
∂
x
2
,
H
z
=
0.
{\displaystyle {\begin{cases}{\begin{array}{ll}{\mathfrak {E}}_{x}=-{\frac {\partial ^{2}\varphi }{\partial x\ \partial z}},&{\mathfrak {H}}_{x}={\frac {v}{c}}{\frac {\partial ^{2}\varphi }{\partial y\ \partial x}},\\\\{\mathfrak {E}}_{y}=-{\frac {\partial ^{2}\varphi }{\partial y\ \partial z}},&{\mathfrak {H}}_{y}=-{\frac {v}{c}}{\frac {\partial ^{2}\varphi }{\partial x^{2}}},\\\\{\mathfrak {E}}_{z}=-{\frac {\partial ^{2}\varphi }{\partial z^{2}}}+{\frac {v^{2}}{c^{2}}}{\frac {\partial ^{2}\varphi }{\partial x^{2}}},&{\mathfrak {H}}_{z}=0.\end{array}}\end{cases}}}
(
E
I
I
)
{\displaystyle ({\mathfrak {E}}_{II})}
{
E
x
=
−
ϰ
2
∂
2
φ
∂
x
∂
z
,
H
x
=
0.
E
y
=
−
∂
2
φ
∂
y
∂
z
,
H
y
=
v
c
∂
2
φ
∂
z
2
,
E
z
=
−
∂
2
φ
∂
z
2
,
H
z
=
−
v
c
∂
2
φ
∂
y
∂
z
.
{\displaystyle {\begin{cases}{\begin{array}{ll}{\mathfrak {E}}_{x}=-\varkappa ^{2}{\frac {\partial ^{2}\varphi }{\partial x\ \partial z}},&{\mathfrak {H}}_{x}=0.\\\\{\mathfrak {E}}_{y}=-{\frac {\partial ^{2}\varphi }{\partial y\ \partial z}},&{\mathfrak {H}}_{y}={\frac {v}{c}}{\frac {\partial ^{2\varphi }}{\partial z^{2}}},\\\\{\mathfrak {E}}_{z}=-{\frac {\partial ^{2}\varphi }{\partial z^{2}}},&{\mathfrak {H}}_{z}=-{\frac {v}{c}}{\frac {\partial ^{2}\varphi }{\partial y\ \partial z}}.\end{array}}\end{cases}}}
Das mit II bezeichnete erhält man aus der Heaviside schen Lösung für eine bewegte Einzelladung durch Superposition.
Wir bilden nun das II entsprechende System für einen magnetischen Dipol, dessen Achse der y -Achse parallel ist, und dessen Moment den Faktor
v
/
c
{\displaystyle v/c}
enthält. Hierfür gewinnen wir
(
v
c
H
I
I
)
{\displaystyle ({\frac {v}{c}}{\mathfrak {H}}_{II})}
{
E
x
=
0
,
H
x
=
v
c
ϰ
2
∂
2
φ
∂
x
∂
y
,
E
y
=
v
2
c
2
∂
2
φ
∂
z
∂
y
,
H
y
=
v
c
∂
2
φ
∂
y
2
,
E
z
=
−
v
2
c
2
∂
2
φ
∂
y
2
,
H
z
=
v
c
∂
2
φ
∂
z
∂
y
.
{\displaystyle {\begin{cases}{\mathfrak {E}}_{x}=0,&{\mathfrak {H}}_{x}={\frac {v}{c}}\varkappa ^{2}{\frac {\partial ^{2}\varphi }{\partial x\ \partial y}},\\\\{\mathfrak {E}}_{y}={\frac {v^{2}}{c^{2}}}{\frac {\partial ^{2}\varphi }{\partial z\ \partial y}},&{\mathfrak {H}}_{y}={\frac {v}{c}}{\frac {\partial ^{2}\varphi }{\partial y^{2}}},\\\\{\mathfrak {E}}_{z}=-{\frac {v^{2}}{c^{2}}}{\frac {\partial ^{2}\varphi }{\partial y^{2}}},&{\mathfrak {H}}_{z}={\frac {v}{c}}{\frac {\partial ^{2}\varphi }{\partial z\ \partial y}}.\end{cases}}}
Addieren wir nun (
E
I
I
{\displaystyle {\mathfrak {E}}_{II}}
) und (
H
I
I
v
c
{\displaystyle {\mathfrak {H}}_{II}{\frac {v}{c}}}
), so gewinnen wir (
E
I
{\displaystyle {\mathfrak {E}}_{I}}
), das nur noch mit dem Faktor
ϰ
2
{\displaystyle \varkappa ^{2}}
multipliziert erscheint.
Bezeichnen wir diese Relation durch die Gleichung
E
I
I
+
v
c
H
I
I
=
ϰ
2
E
I
,
{\displaystyle {\mathfrak {E}}_{II}+{\frac {v}{c}}{\mathfrak {H}}_{II}=\varkappa ^{2}{\mathfrak {E}}_{I},}
so können wir eine entsprechende herleiten
v
c
H
I
I
+
v
2
c
2
E
I
I
=
v
c
ϰ
2
H
I
,
{\displaystyle {\frac {v}{c}}{\mathfrak {H}}_{II}+{\frac {v^{2}}{c^{2}}}{\mathfrak {E}}_{II}={\frac {v}{c}}\varkappa ^{2}{\mathfrak {H}}_{I},}
woraus
E
I
I
(
1
−
v
2
c
2
)
=
ϰ
2
(
E
I
−
v
c
H
I
)
,
{\displaystyle {\mathfrak {E}}_{II}\left(1-{\frac {v^{2}}{c^{2}}}\right)=\varkappa ^{2}\left({\mathfrak {E}}_{I}-{\frac {v}{c}}{\mathfrak {H}}_{I}\right),}
E
I
I
=
E
I
−
v
c
H
I
{\displaystyle {\mathfrak {E}}_{II}={\mathfrak {E}}_{I}-{\frac {v}{c}}{\mathfrak {H}}_{I}}
folgt.